Poj 3468-A Simple Problem with Integers 线段树,树状数组

题目：http://poj.org/problem?id=3468

 

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 85851		Accepted: 26685
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

 

题意:给一串数字，每次区间加上一个数，或询问区间和。

 

题解:

线段树水过，打个lazy标记就好。。。

线段树程序：

 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define LL long long
 #define MAXN 100010
 struct node
 {
     LL left,right,sum,tag;
 }tree[MAXN*];
 LL A[MAXN];
 LL read()
 {
     LL s=,fh=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}
     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}
     return s*fh;
 }
 void Pushup(LL k)
 {
     tree[k].sum=tree[k*].sum+tree[k*+].sum;
 }
 void Pushdown(LL k,LL l,LL r)
 {
     if(tree[k].tag!=)
     {
         tree[k*].tag+=tree[k].tag;
         tree[k*+].tag+=tree[k].tag;
         LL mid=(l+r)/;
         tree[k*].sum+=tree[k].tag*(mid-l+);
         tree[k*+].sum+=tree[k].tag*(r-mid);
         tree[k].tag=;
     }
 }
 void Build(LL k,LL l,LL r)
 {
     tree[k].left=l;tree[k].right=r;
     if(l==r)
     {
         tree[k].sum=A[l];
         return;
     }
     LL mid=(l+r)/;
     Build(k*,l,mid);Build(k*+,mid+,r);
     Pushup(k);
 }
 void Add(LL k,LL ql,LL qr,LL add)
 {
     //Pushdown(k,tree[k].left,tree[k].right);
     if(ql<=tree[k].left&&tree[k].right<=qr){tree[k].tag+=add;tree[k].sum+=(tree[k].right-tree[k].left+)*add;return;}
     Pushdown(k,tree[k].left,tree[k].right);
     LL mid=(tree[k].left+tree[k].right)/;
     if(qr<=mid)Add(k*,ql,qr,add);
     else if(ql>mid)Add(k*+,ql,qr,add);
     else {Add(k*,ql,mid,add);Add(k*+,mid+,qr,add);}
     Pushup(k);
 }
 LL Sum(LL k,LL ql,LL qr)
 {
     //Pushdown(k,tree[k].left,tree[k].right);
     if(ql<=tree[k].left&&tree[k].right<=qr)return tree[k].sum;
     Pushdown(k,tree[k].left,tree[k].right);
     LL mid=(tree[k].left+tree[k].right)/;
     if(qr<=mid)return Sum(k*,ql,qr);
     else if(ql>mid)return Sum(k*+,ql,qr);
     else return Sum(k*,ql,mid)+Sum(k*+,mid+,qr);
 }
 int main()
 {
     LL N,Q,i,a,b,c;
     char fh[];
     N=read();Q=read();
     for(i=;i<=N;i++)A[i]=read();
     Build(,,N);
     for(i=;i<=Q;i++)
     {
         scanf("\n%s",fh);
         if(fh[]=='Q')
         {
             a=read();b=read();
             printf("%lld\n",Sum(,a,b));
         }
         else
         {
             a=read();b=read();c=read();
             Add(,a,b,c);
         }
     }
     return ;
 }

树状数组不会。。。

先附上fhq神犇的题解：

上次NOI集训的时候，一位福建女神牛和我探讨过这题能不能用BIT，我当时
的答复是可以，因为“扩展树状数组”这个东西理论上可以实现一般线段树
可以实现的东西，且空间上的常数好一点。但是对于“扩展树状数组”，这
个东西是我一时兴起想到的玩意，没有进行更多的研究，没查到任何的资料
，更没有想过如何把线段树著名的lazy思想照搬上去，以及动态开内存的解
决方案。

权衡利弊，我想了一个使用两棵BIT的替代方法来解决这题，并且成功地将
内存使用做到了1728K。这恐怕是带标记的扩展树状数组达不到的。

记录两个BIT，
设数列为A[i],BIT1的每个元素B1[i]=A[i]*i,
BIT2的每个元素B2[i]=A[i]

则：sum{A[i]|i<=a}=sum{B1[i]|i<=a}+(sum{B2[i]|1<=i<=N}-sum{B2[i]|i<=a})*a
sum{A[i]|a<=i<=b}=sum{A[i]|i<=b}-sum{A[i]|i<a}
这样就十分巧妙的解决了！

关键代码：
int N;
struct BIT{
	long long a[NMax];
	void ins(int x,long long k){
		for (;x<N;x+=((x+1)&-(x+1)))a[x]+=k;
	}
	long long ask(int x){
		long long ret;
		for (ret=0;x>=0;x-=((x+1)&-(x+1)))ret+=a[x];
		return ret;
	}
}B1,B2;
long long B2S;
void Add(int a,long long x){
	//printf("Add %d %I64d\n",a,x);
	B1.ins(a,x*((long long)a+1));
	B2.ins(a,x);B2S+=x;
}
void Add(int a,int b,long long x){
	Add(b,x);
	if (a)Add(a-1,-x);
}
long long Ask(int a){
	//printf("Ask %d\n",a);
	long long ret=B1.ask(a)+(B2S-B2.ask(a))*(long long)(a+1);
	//printf("=%I64d\n",ret);
	return ret;
}
long long Ask(int a,int b){
	long long ret;
	ret=Ask(b);
	if (a)ret-=Ask(a-1);
	return ret;
}

填坑中。。。。。。